Nested intervals theorem - a special case on open intervals

For all $n,$ $[a_n, b_n] \supset (a_n, b_n) \supset [a_{n+1}, b_{n+1}],$ therefore $([a_n, b_n])_{n\geqslant1}$ is a nested sequence of closed intervals, therefore $$ \bigcap_{n=1}^\infty(a_n, b_n) \supseteq \bigcap_{n=1}^\infty[a_{n+1}, b_{n+1}] \ne \varnothing. $$

Hints: Let $A=\sup_n a_n =\lim a_n$. Since $(a_n)$ is strictly increasing it follows that $a_n <A$ for all $n$. Now $a_{n+k} <a_n <b_n$ for all $n,k$ so $A=\lim_k a_{n+k} \leq b_n$ for all $n$. Suppose $A=b_n$ for some $n$. Then $b_{n+1}<b_n=A(=\lim a_n)$. This implies that $b_{n+1}<a_m$ for all $m$ sufficiently large. But then $a_m <b_m <b_{n+1} <a_m$ when $m$ is large enough. This contradiction shows that $A<b_n$ for al $n$. Hence $A \in (a_n,b_n)$ for all $n$.

Our assumption is that $$a_{n} < a_{n+1},a_n<b_n,b_n>b_{n+1}\tag{1}$$ Then by density of reals we can find numbers $a'_n, b'_n$ such that $$a_n<a'_n<a_{n+1},b_n>b'_n>b_{n+1}\tag{2}$$ Then the above inequalities lead to $$a'_n<a'_{n+1},a'_n<b'_n,b'_n>b'_{n+1}\tag{3}$$ Thus the sequence of closed intervals $[a'_n, b'_n] $ is nested and has a non-empty intersection. So there is a real number $x$ such that $x\in[a'_n, b'_n]\subset (a_n, b_n) $ for all $n$. So the intervals $(a_n, b_n) $ have a non-empty intersection.